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Introduction to the square root of a 2 by 2 matrixYue Kwok ChoyThe square root of a 2 by 2 matrix A is another 2 by 2 matrix B such that A=B2, where B2 stands for the matrix product of B with itself. We write A12=B. In general, there can be no, two, four or even an infinite number of square root matrices.For example: 12341234=7101522If we take A=7101522 , then A12=710152212=1234Note that B=1234 is one of the square roots of A=7101522 . There may be another root(s) for A.Questions (See answers at the end of this file)1.Evaluate : 11111111. Hence find two square roots of (a) 2222 (b) 1111 .2.Find four values of 400912 . Hence find four values of a00b12 , where a,b≥0 .3.List all possible matrices A=abcd , where a, b, c and d can be either 0 or 1, such that A12 does not exist.(A)Solving equations methodSuppose we like to find the square root of A=1225. So we write:abcdabcd=1224?a2+bcba+dca+dcb+d2=1225We get four equations:a2+bc=1 …(1)ba+d=2 …(2)ca+d=2 …(3)cb+d2=5 …(4)2-3, b-ca+d=0Since a+d≠0 (otherwise (2) or (3) give absurdity), b-c=0, b=c.Hence we get: a2+b2=1 …(5)ba+d=2 …(6)b2+d2=5 …(7)(6)2, b2a+d2=4 , b2=4a+d2 …(8) 8↓5, a2+4a+d2=1, …(9)7-5,d2-a2=4 …(10) d-a= 4a+d …(11)(a+d≠0)(11)↓(9), a2+d-aa+d=1, a3+a2d+d-a=a+d, a3+a2d-2a=0aa2+ad-2=0∴a=0 or d=2-a2a …(12)12↓10, a=0, d=±2 or a=±12, d=±32The rest can be solved easily, we therefore have:122512=0±1±1±2, ±12±12±12±323.Prove that 010012 does not exist.List, without prove, all possible matrices A=abcd , where a, b, c and d can be either 0 or 1, such that A12 has no real solution(s).4.Use algebraic method to find 120112.5.Check that the square root of the identity matrix is given by:100112=±d1-d2cc?d,±dc1-d2c?d ,where ±±100?1,±10c?1,±1c0?1 are limiting cases.6.(a)Let A=1224, find A and A2. Hence find A12 .(b)A=abcd, where A=0. Given that trA=a+d>0.Show that A2=trAA . Hence find A12 .(B)Diagonalization of MatrrixSolving equation method in finding the square root of a matrix may not be easy. It involves solving four non-linear equations with four unknowns. You may try this: 4112123412, and soon may give up. We note that the square root of a diagonal matrix can be found easily: a00b12=a00b,-a00b,a00-b,-a00-b.If a matrix is NOT a diagonal matrix, we devise a method called diagonalization to help us.We proceed with the finding of the eigenvalue(s) and eigenvector(s) of A .A real number is said to be an eigenvalue of a matrix A if there exists a non-zero column vector v such that Av=λv or A-λIv=0 .We like to find 3324485712(1)EigenvaluesA=33244857, v=xy, A-λIv=33-λ244857-λxy=0Now, A-λIv=0 has non-zero solution, A-λI=033-λ244857-λ=033-λ57-λ-48×24=0λ2-90λ+729=0λ-81 λ-9=0∴λ=9 or λ=81 , and these are the eigenvalues.(2)EigenvectorsWe usually would like to find the eigenvector corresponding to each eigenvalue.The process is called normalization.For λ1=9,33244857x1y1=9x1y1 33x1+24y1=9x148x1+57y1=9y1Choose for convenience24x1+24y1=0 x1+y1=0 x1y1=1-1 , which is a eigenvector.For λ2=81,33244857x2y2=81x2y2 33x2+24y2=81x248x1+57y1=81y1Choose -48x1+24y1=0 2x1-y1=0 x2y2=12 , which is another eigenvector.(3)Diagonalization of matrixWe place two eigenvectors together. Let P=x1y1x2y2=1-112Consider another matrix:B=P-1AP=1-112-1332448571-112=1321-11332448571-112=90081 At last we get a matrix B which is diagonal with eigenvalues as entries in the main diagonal.B12=9008112=3009,-300-9,300-9,-3009 Note:The diagonalization of a matrix may not be a simple subject since A-λI=0 may have equal roots or even complex roots. Although most matrices are not diagonal, they can be diagonalized. Not all square matrices can be diagonalised.A thorough study of diagonalization of a matrix is not discussed here.(4)Finding the square root of the original matrix ASince B=P-1AP, we have A=PBP-1PB12P-1PB12P-1=PB12P-1PB12P-1=PB12B12P-1=PBP-1=ASo PB12P-12=A(a)A12=PB12P-1=1-11230091-112-1=1-11230091321-11=3-39181321-11=1-13621-11=5427(b)A12=PB12P-1=1-112-30091-112-1=1-112-30091321-11=-339181321-11=-113621-11=1485∴3324485712=5247,-5-2-4-7,1485,-1-4-8-5Questions7.Real matrix may have irrational square roots. Check by multipication:120312=13-103,1-3-10-3,-13+103,-1-3+10-38.A simple real matrix may have complex square roots. Check by multipication:011012=121+i1-i1-i1+i,121-i1+i1+i1-i,12-1-i-1+i-1+i-1-i,12-1+i-1-i-1-i-1+i9.A matrix can have both integral and fractional square roots. Use Diagonalization Method to show: 04-1512=±2-41-3,±2343-1373Answers1.11111111=2222 , Hence:(a)222212=1111, -1-1-1-1(b)11111111=2222=21111?1111=±121111±121111111112=±121111=12121212, -12-12-12-122.400912=2003,-2003,200-3,-200-3a00b12=a00b,-a00b,a00-b,-a00-b3.abcdabcd=0100?a2+bcba+dca+dcb+d2=0100a2+bc=0=cb+d2?a=±d(a) Since ba+d=1, a+d≠0, and so a=d≠0.(b) Lastly, since ca+d=0?c=0 and so a2+bc=a2=0, contradicts with (a).010012, 001012 does not exist011112111012 has no real solution(s).4.abcdabcd=1201?a2+bcba+dca+dcb+d2=1201We get four equations:a2+bc=1 …(1)ba+d=2 …(2)ca+d=0 …(3)cb+d2=1 …(4)From (3), c=0 …(5) (a+d=0 gives contradiction in (2))From (1), a2=1, a=±1.From (3),d2=1, d=±1Hence, 120112=1101,-1-10-16.(a)A=1224, A=0 , A2=12241224=5101020=51224=5A. Hence A=15A2=15A2 . A12=15A=151224=15252545 .(b)A=abcd, A=ad-bc=0?ad=bc …(1) A2=abcdabcd=a2+bcba+dca+dcb+d2=a2+adba+dca+dad+d2 , by (1).=aa+dba+dca+dda+d=a+dabcd=trAA∴A=1trAA2=1trAA2, where trA>0.Hence, A12=1trAA .9.Eigenvalues : 4, 1Eigenvectors: 11,41P=x1y1x2y2=1141, P-1=13-114-1 , B= 4001B12=400112=2001,-200-1,200-1,-2001(a)A12=PB12P-1=114120011141-1=1141200113-114-1=2343-1373(b) A12=PB12P-1=1141-20011141-1=1141-200113-114-1=2-41-3∴04-1512=±2-41-3,±2343-1373 ................
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